In the previous chapter we dealt with motion in the plane of symmetry. Motion of the plane of symmetry will now be considered. This motion consists of transition in the у direction called sideslip; rotation about the x-axis referred to as rolling; and rotation about the z-axis, or yawing. To begin, the equations of motion will once again be derived, but in a somewhat more basic and complete manner than that which was followed in Chapter Nine. The following derivation closely parallels a similar development presented by Seckel (Ref. 10.4).

Figure 10.1 illustrates a particle of mass of the airplane located at the point x, y, z, in the moving-body axis system. As shown, the axis system is translating with instantaneous velocity components of U, V, and W in the x, y, and z directions while rotating about these axes at angular rates of P, Q, and R.

The linear velocity components of AM in the x, y, and z directions are, obviously,

x = U + Qz – Ry у = V-Pz + Rx z = W + Py – Qx

The accelerations are obtained directly by differentiating the above velocities.

x = U + Qz + Qz – Ry – Ry у – V – Pz – Pz + Rx + Rx z = W + Py + Py – Qx – Qx

Figure 10.1 Motion of a particle of mass in a moving reference system.

x, y, and z are obtained from Equations 10.1, so Equation 10.2 becomes:

x = U + QW + QPy – Q2x – RV + RPz – R2x + Qz-Ry (10.3a)

у = V – PW – P2y – PQx+ RU + RQz – R2y – Pz + Rx (10.3b)

z = W + PV – P2z + PRx – QU – Q2z + QRy + Py – Qx (10.3c)

It is convenient at this point to use the concept of inertia forces and moments. Briefly, this concept allows one to treat a dynamic system as a static one by employing pseudoforces acting on the system equal in mag­nitude to the product of each mass and its acceleration. The forces are directed opposite to the accelerations. As an example, consider Figure 10.2. The dynamic equation of motion is, obviously,

F = Mx (10.4)

Now, however, add a force on the mass opposite in direction to x equal to Mx. As a problem in statics, the sum of the forces on the mass equals zero.

F – Mx = 0

This, of course, is equal to Equation 10.4.

The inertia forces on the mass element can be written as

FXi = – mx (10.5a)

Fy. = – my (10.5b)

Fz. = – mz (10.5c)

Figure 10.2 Concept of inertia force.

with x, y, and z given by Equation 10.3. These inertia forces give rise to inertia moments. The moment arms can be seen in Figure 10.1. For example, an x force on Am gives rise to a moment about the у-axis equal to FXZ. Therefore, the three inertia moments become:

Mx. = Fz. y – Fyz (10.6a)

Mn = FXiz – Fzx (10.66)

MZI = Fyx – Fx. y (10.6c)

This can be written in vector form as

M, = Rx F,

The total inertia forces and moments acting on the airplane are obtained by summing Equations 10.5 and 10.6 over the total airplane mass. In evaluat­ing these sums, the following sums vanish since the origin of the coordinate system is at the center of gravity

2 A mx = 0 2 bmy = 0 2Amz = 0

Also, because of symmetry,

2 A mxy = 0

Not included in these equations are angular momentum vectors that may be present because of rotors or other rotating components. If h denotes such an angular momentum vector, then ы x h must be added to the above momentum equations. Thus Equation 10.8d, 10.8c, and 10.8/ becomes,

L = IXP – IXZR + Qhz – Rhy

M = IyQ – Phy + Rhx N = IZR – IXZP + Phy – Rhz

Equations 10.8a, 10.8c, and 10.8c were derived previously in Chapter Nine and were used to examine longitudinal dynamic stability and control. Equa­tions 10.8b, 10.8d, and 10.8/ will be used for the analysis of lateral-directional stability and control.

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